As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force acting on it is equal to (41 - 9j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?

v_f = 10.38 m / s

Explanation:

For this exercise we can use the relationship between work and kinetic energy

          W = ΔK

note that the two quantities are scalars

Work is defined by the relation

          W = F. Δx

the bold are vectors.  The displacement is

          Δx = r_f -r₀

          Δx = (11.6 i - 2j) - (4.4 i + 5j)

          Δx = (7.2 i - 7 j) m

          W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

           i.i = j.j = 1

           i.j = 0

           W = 4  7.2 + 9  7

           W = 91.8 J

the initial kinetic energy is

           Ko = ½ m vo²

           Ko = ½ 2.0 4.0²

           Ko = 16 J

we substitute in the initial equation

          W = K_f - K₀

          K_f = W + K₀

          ½ m v_f² = W + K₀

          v_f² = 2 / m (W + K₀)

          v_f² = 2/2 (91.8 + 16)

           v_f = √107.8

          v_f = 10.38 m / s

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

it becomes positively charged by charging it